Lewis structure of XeO₃
26 valence electrons. Three Xe=O double bonds, with one lone pair on Xe, so Xe has 14 electrons around it (an expanded octet).
Lewis structure
26 valence electronsFormal charges: none
Why this shape
- Xe has 8 valence electrons. It uses six in three double bonds to O, so two are left as one lone pair.
- Steric number = three sigma bonds + one lone pair = 4, so Xe is sp³ hybridised. A double or triple bond counts once: only its sigma bond uses a hybrid orbital.
- Four electron pairs point to the corners of a tetrahedron (tetrahedral electron geometry). With one lone pair in place of atoms, the atoms make a trigonal pyramidal shape.
- Repulsion falls in the order lone pair to lone pair > lone pair to bond pair > bond pair to bond pair. The lone pair pushes the bond pairs closer, so the angle is about 103°, less than the ideal 109.5°.
- The bond dipoles do not cancel in the trigonal pyramidal shape, so XeO₃ is polar.
How to draw it
- Count the valence electrons: Xe 8 + 3 × O 6 = 26 valence electrons, 13 pairs.
- Put Xe in the middle and join it to the three outer atoms with single bonds.
- Give each outer atom lone pairs until it has an octet (H needs only two electrons), then put any electrons left over on the central atom.
- Where an atom is still short of an octet, turn a lone pair of its neighbour into a shared pair: that makes a double or triple bond.
- Result: three Xe=O double bonds, with one lone pair on Xe, so Xe has 14 electrons around it (an expanded octet).
- Formal charge = valence electrons − lone pair electrons − number of bonds. Here: none.
Class 11, Chapter 4 Chemical Bonding and Molecular Structure.
Lewis structure of XeO3: quick answers
How many valence electrons are in XeO3?
Xe 8 + 3 × O 6 = 26 valence electrons, 13 pairs.
What are the formal charges in the Lewis structure of XeO3?
Every atom has a formal charge of zero.
How many lone pairs are in the Lewis structure of XeO3?
7 in all, 1 of them on the central Xe.